ln(x)
----
..x
= ln(x) / x
Let y = ln(x) / x
Differentiate both sides with respect to x,
dy/dx = {x*[1/x] -ln(x) *[1]} / x^2
dy/dx = {1 -ln(x)} / x^2
That is it.
We used
d/dx (u/v) = {v*du -u*dv} / v^2
x = e?
if so, then
ln(x) / x
= ln(e) / e
= 1/e
And differentiating that will give you zero because 1/e is a constant.
---------
If you substitute e for x in the derivative dy/dx = {1 -ln(x)} / x^2,
dy/dx = {1 -ln(e)} / e^2
dy/dx = {1 -1} / e^2
dy/dx = {0} / e^2
dy/dx = 0 ------------------still zero.